Question

A uniform rod of mass $ m $ and length $ l $ can rotate freely on a smooth horizontal plane about a vertical axis hinged at point H. A point mass having same mass $ m $ coming with an initial speed $ u $ perpendicular to the rod, strikes the rod inelastically at its free end. Find out the angular velocity of the rod just after collision?

Answer 1

Hint: To solve this question we need to use the theorem of conservation of angular momentum. For that we have to calculate the moment of inertia of the system about the vertical axis passing through H.

Formula used: The formula which is used to solve this question is given by
 $ L = I\omega $, here $ L $ is the angular momentum of a body about an axis which has a moment of inertia of $ I $ about the same axis and is rotating with an angular velocity of $ \omega $.

Complete step by step answer
During the collision of the point mass with the rod, there is no external force acting on the system of the rod and the point mass, which can produce external torque about the point H. So the angular momentum of the system about the point H will be conserved before and after the collision.
Initial angular momentum:
As the rod is initially at rest, so its initial angular momentum about the point H is equal to zero.
Now, the point mass is moving perpendicular to the rod. So its angular momentum about the point H is equal to $ mvl $. So the total initial angular momentum of the system is
 $ {L_i} = 0 + mul = mul $....................................(1)
Final angular momentum:
As the collision is inelastic, the point mass gets attached to the rod and they both start rotating with the same angular velocity, say $ \omega $.
We know that the angular momentum of a rotating body is given by
 $ L = I\omega $..................................(2)
Now, we have to calculate the moment of inertia of the rod and the point mass about the point H.
The moment of inertia of the point mass about the vertical axis passing through H is
 $ {I_1} = m{l^2} $
Also the moment of inertia of the rod is
 $ {I_2} = \dfrac{{m{l^2}}}{3} $
So the net moment of inertia of the system is
 $ I = {I_1} + {I_2} $
 $ \Rightarrow I = m{l^2} + \dfrac{{m{l^2}}}{3} $
On solving we get
 $ I = \dfrac{{4m{l^2}}}{3} $
From (2) the final angular momentum of the system is given by
 $ {L_f} = \dfrac{{4m{l^2}}}{3}\omega $.....................................(3)
As the initial and the final angular momentum are equal, so from (1) and (3) we have
 $ \dfrac{{4m{l^2}}}{3}\omega = mul $
Cancelling $ ml $ from both the sides we have
 $ \dfrac{{4l}}{3}\omega = u $
 $ \Rightarrow \omega = \dfrac{{3u}}{{4l}} $
Hence, the angular velocity of the rod just after collision is equal to $ \dfrac{{3u}}{{4l}} $.

Note
We should not try to conserve the kinetic energy of the system to get the final answer. This is because the collision is inelastic, so some of the kinetic energy of the point mass will be lost in the form of heat, deformations etc. Hence, the final kinetic energy of the system will be less than the initial kinetic energy.